ach linear equation given (ii) y = y = x – (iv) y = 4x – 5 (vi) 2x – 3y = 4 plz give me equation it’s u Question ach linear equation given(ii) y =y = x –(iv) y = 4x –5(vi) 2x – 3y = 4 plz give me equation it’s urgent in progress 0 Math Maya 7 hours 2021-11-25T04:12:36+00:00 2021-11-25T04:12:36+00:00 1 Answer 0 views 0

## Answers ( )

Answer:(i) x + y =5 and 2x –3y = 4

By elimination method

x + y =5 … (i)

2x –3y = 4 … (ii)

Multiplying equation (i) by (ii), we get

2x + 2y = 10 … (iii)

2x –3y = 4 … (ii)

Subtracting equation (ii) from equation (iii), we get

5y = 6

y = 6/5

Putting the value in equation (i), we get

x = 5 – (6/5) = 19/5

Hence, x = 19/5 and y = 6/5

By substitution methodx + y = 5 … (i)

Subtracting y both side, we get

x = 5 – y … (iv)

Putting the value of x in equation (ii) we get

2(5 – y) – 3y = 4

-5y = – 6

y = -6/-5 = 6/5

Putting the value of y in equation (iv) we get

x = 5 – 6/5

x = 19/5

Hence, x = 19/5 and y = 6/5 again

(ii) 3x + 4y = 10 and 2x – 2y = 2

By elimination method

3x + 4y = 10 …. (i)

2x – 2y = 2 … (ii)

Multiplying equation (ii) by 2, we get

4x – 4y = 4 … (iii)

3x + 4y = 10 … (i)

Adding equation (i) and (iii), we get

7x + 0 = 14

Dividing both side by 7, we get

x = 14/7 = 2

Putting in equation (i), we get

3x + 4y = 10

3(2) + 4y = 10

6 + 4y = 10

4y = 10 – 6

4y = 4

y = 4/4 = 1

Hence, answer is x = 2, y = 1

By substitution method

3x + 4y = 10 … (i)

Subtract 3x both side, we get

4y = 10 – 3x

Divide by 4 we get

y = (10 – 3x )/4

Putting this value in equation (ii), we get

2x – 2y = 2 … (i)

2x – 2(10 – 3x )/4) = 2

Multiply by 4 we get

8x – 2(10 – 3x) = 8

8x – 20 + 6x = 8

14x = 28

x = 28/14 = 2

y = (10 – 3x)/4

y = 4/4 = 1

Hence, answer is x = 2, y = 1 again.

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

By elimination method

3x – 5y – 4 = 0

3x – 5y = 4 …(i)

9x = 2y + 7

9x – 2y = 7 … (ii)

Multiplying equation (i) by 3, we get

9 x – 15 y = 11 … (iii)

9x – 2y = 7 … (ii)

Subtracting equation (ii) from equation (iii), we get

-13y = 5

y = -5/13

Putting value in equation (i), we get

3x – 5y = 4 … (i)

3x – 5(-5/13) = 4

Multiplying by 13 we get

39x + 25 = 52

39x = 27

x =27/39 = 9/13

Hence our answer is x = 9/13 and y = – 5/13

By substitution method

3x – 5y = 4 … (i)

Adding 5y both side we get

3x = 4 + 5y

Dividing by 3 we get

x = (4 + 5y )/3 … (iv)

Putting this value in equation (ii) we get

9x – 2y = 7 … (ii)

9 ((4 + 5y )/3) – 2y = 7

Solve it we get

3(4 + 5y ) – 2y = 7

12 + 15y – 2y = 7

13y = – 5

y = -5/13

x = 4 + 5 ( -5/13)/ 3

= 4 – 25/13 / 3

= 4 × 13 – 25/13 / 3

= 27/13×3

= 27/39

= 9/13

Hence we get x = 9/13 and y = – 5/13 again.

(iv) x/2 + 2y/3 = – 1 and x – y/3 = 3

By elimination method

x/2 + 2y/3 = -1 … (i)

x – y/3 = 3 … (ii)

Multiplying equation (i) by 2, we get

x + 4y/3 = – 2 … (iii)

x – y/3 = 3 … (ii)

Subtracting equation (ii) from equation (iii), we get

5y/3 = -5

Dividing by 5 and multiplying by 3, we get

y = -15/5

y = – 3

Putting this value in equation (ii), we get

x – y/3 = 3 … (ii)

x – (-3)/3 = 3

x + 1 = 3

x = 2

Hence our answer is x = 2 and y = −3.

By substitution method

x – y/3 = 3 … (ii)

Add y/3 both side, we get

x = 3 + y/3 … (iv)

Putting this value in equation (i) we get

x/2 + 2y/3 = – 1 … (i)

(3+ y/3)/2 + 2y/3 = -1

3/2 + y/6 + 2y/3 = – 1

Multiplying by 6, we get

9 + y + 4y = – 6

5y = -15

y = – 3

Hence our answer is x = 2 and y = −3.